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15 tháng 7 2015

\(C=0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)

     \(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)

    \(=\frac{15+10+12+5}{30}+\frac{25-4}{35}+\frac{1}{41}\)

    \(=\frac{7}{5}+\frac{3}{5}+\frac{1}{41}\)

   \(=2+\frac{1}{41}=\frac{83}{41}\)

 

\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

     \(=\left(\frac{1}{90}-\frac{1}{30}-\frac{1}{6}-\frac{1}{2}\right)+\left(-\frac{1}{72}-\frac{1}{12}\right)-\frac{1}{56}-\frac{1}{42}\)

    \(=\frac{1-2-15-45}{90}+\frac{-1-6}{72}-\frac{1}{56}-\frac{1}{42}\)

    \(=-\frac{61}{90}-\frac{7}{72}-\frac{1}{56}-\frac{1}{42}\)

    \(=\frac{-1708-245-45-60}{2520}\)

    \(=-\frac{49}{60}\)

Nghịch đảo của C là \(\frac{41}{83}\), nghịch đảo của D là \(-\frac{60}{49}\)

\(\frac{41}{83}\cdot\left(-\frac{60}{49}\right)=-\frac{2460}{4067}\)

    

30 tháng 6 2016
  1. C = \(\frac{83}{41}\)
  2. D = \(\frac{-79}{90}\)
20 tháng 4 2016

C=0.5+1/3+0.4+5/7+1/6-4/35+1/41

C=1/2+1/3+2/5+5/7+1/6-4/35+1/41

C=(1/2+1/6+1/3)+(2/5+5/7-4/35)+1/41

C=1+1-1/41

C=2-1/41

=>C=81/41

D=1/90-1/72-1/56-1/42-1/30-1/20--1/12-1/6-1/2

=> D=1/9*10-1/8*9-1/7*8-1/6*7-1/5*6-1/4*5-1/3*4-1/2*3-1/1*2

=>D=-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=>D=-(1-1/10)

=>D=-9/10

ai k mh mh k lại

5 tháng 3 2018

C.D=18/10 nha

31 tháng 7 2016

a) \(\frac{14}{7}+\frac{5}{9}+\frac{4}{9}-2\)

\(=2\frac{5}{9}+\frac{4}{9}-2\)

\(=3-2=1\)

b) \(\frac{31}{25}+\left(\frac{15}{16}-\frac{6}{15}\right)+\frac{1}{16}\)

\(=\frac{31}{25}+\frac{43}{80}+\frac{1}{16}\)

\(=1\frac{311}{400}+\frac{1}{16}\)

\(=1\frac{21}{25}\)

15 tháng 10 2018

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

14 tháng 4 2021

−190−172−156−142−130−120−112−16−12

=−190−(12+16+112+120+130+142+156+172)

=−190−(11.2+12.3+13.4+14.5+15.6+16.7+17.8+18.9)

=−190−(1−12+12−13+13−14+14−15+15−16+16−17+17−18+18−19)

=−190−(1−19)

=−190−89

8 tháng 3 2019

\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)

\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)

\(\text{c,d cơ bản tự làm nha }\)

8 tháng 3 2019

A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9

=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9

=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9

=>1/2-1/9=>9/18-2/18=>7/18

Vậy A= 7/18

3 tháng 2 2022

a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)

b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)

=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).

d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).

e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)